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Proof of convergence and divergence

  1. Oct 1, 2006 #1
    Prove that if [tex] a_{n} > 0 [/tex] and [tex] \sum a_{n} [/tex] converges, then [tex] \sum a_{n}^{2} [/tex] also converges.

    So if [tex] \sum a_{n} [/tex] converges, this means that [tex] \lim_{n\rightarrow \infty} a_{n} = 0 [/tex]. Ok, so from this part how do I get to this step: there exists an [tex] N [/tex] such that [tex]| a_{n} - 0 | < 1 [/tex] for all [tex] n > N \rightarrow 0\leq a_{n} < 1 [/tex]. Thus [tex] 0\leq a_{n}^{2} \leq a_{n} [/tex]. How did we choose [tex] |a_{n} - 0| < 1 [/tex]?
     
  2. jcsd
  3. Oct 1, 2006 #2
    nvm got it. just arbritrary number.
     
  4. Oct 1, 2006 #3

    AKG

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    Are you sure you got it? There's not much in your first post that looks right, and "just arbitrary number" doesn't seem to make any sense, or have much at all to do with proving the desired claim.
     
  5. Oct 2, 2006 #4
    It loox fine to me.
    [tex] \sum a_{n} [/tex] converges ==> [tex] \lim_{n\rightarrow \infty} a_{n} = 0 [/tex] ==> there exists an [tex] N [/tex]
    such that for all [tex] n > N \rightarrow 0\leq a_{n} < 1 [/tex]
     
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